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# Whoopsies posted this off topic

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A spherical balloon is being inflated. Its volume is increasing at the rate of 2cm^3 per second. Find the rate in cm/sec at which the radius of the balloon is increasing when the radius of the balloon is 4cm.

Any help would be appreciated :)

Chelle

Aug 3, 2019

#1
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Hi Chelle,

A spherical balloon is being inflated. Its volume is increasing at the rate of 2cm^3 per second. Find the rate in cm/sec at which the radius of the balloon is increasing when the radius of the balloon is 4cm.

$$V=\frac{4}{3}\pi r^3\\ \frac{dV}{dr}=4\pi r^2 \\~\\ \frac{dV}{dt}=2,\qquad \text{Find \frac{dr}{dt} when r=4}\\~\\ \frac{dr}{dt}=\frac{dr}{dV}\cdot \frac{dV}{dt}=\frac{1}{4\pi r^2}\cdot 2=\frac{1}{2\pi r^2}\\~\\ so\\ \text{When r=4 the rate at which the radius of the balloon is increasing is }\quad \dfrac{1}{32\pi }\;\;cm/sec$$

Aug 3, 2019
#3
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A spherical balloon is being inflated. Its volume is increasing at the rate of 2cm^3 per second. Find the rate in cm/sec at which the radius of the balloon is increasing when the radius of the balloon is 4cm.

$$r:\frac{4}{3}\pi r^3=\Delta r:\Delta V\\ 1:\frac{4}{3}\pi r^2=\Delta r:\Delta V\\ 1:(\frac{4}{3}\cdot \pi \cdot16cm^2)=\Delta r:(2cm^3/s)\\ \Delta r=\frac{2cm^3\cdot 3}{s\cdot4\pi \cdot 16cm^2}$$

$$\Delta r=0,0298cm/s$$ !

Aug 3, 2019
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Shouldn't second line be

4 pi r^2     (as the 3 brought down from the exponent cancels the 3 in the denominator)  ?

Guest Aug 3, 2019